$4-3+1-2-4+3-7+8+10$
$5\left(2x-4\right)\le4x+10$
$f\left(x\right)=\left(1+\frac{1}{x}\right)^{2x}$
$5-3\left(4-11\right)+10\left(4-5\right)-15$
$\{[\left(-15\right)^{2}]^{0}\}^{5}$
$\lim_{x\to0}\left(\frac{2sin-0.001}{-0.001}\right)$
$x^2+9x+9$
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